The derivative of x raised to the power of x with respect to $x$ is written in the following mathematical form.

$\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$

Let’s get started to learn how to find the derivative of this exponential power function with respect to variable $x$.

In differential calculus, there are formulas for finding the derivative of both exponential function and power function but there is no derivative rule to find the differentiation of a special function, which is formed by the combination of both exponential and power functions and it creates the problem while evaluating the derivative of this special function. However, it can be overcome by eliminating the exponential notation from this function.

Let’s assume that $y$ $\,=\,$ $x^{\displaystyle x}$

Now, take natural logarithm to expressions on both sides of the equation.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $\log_{e}{\big(x^{\displaystyle x}\big)}$

Use the power rule of logarithms for eliminating the exponential form the exponential power function.

$\implies$ $\log_{e}{(y)}$ $\,=\,$ $x \times \log_{e}{(x)}$

Now, differentiate the expressions on both sides of the logarithmic equation with respect to $x$.

$\implies$ $\dfrac{d}{dx}{(\log_{e}{y})}$ $\,=\,$ $\dfrac{d}{dx}{(x \times \log_{e}{x})}$

On left-hand side of the equation, $y$ is a variable in the logarithmic expression but it should be differentiated with respect to $x$. The differentiation of natural logarithm of $y$ with respect to $x$ can be evaluated by chain rule.

On right-hand side of the equation, the variable $x$ and the natural logarithm of $x$ are multiplied and the derivative of their product has to calculate. It can be calculated by the product rule of derivatives.

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $x \times \dfrac{d}{dx}{(\log_{e}{x})}$ $+$ $\log_{e}{x} \times \dfrac{d}{dx}{(x)}$

As per the derivative rule of logarithms, the derivative of natural logarithm of $x$ with respect to $x$ is the reciprocal of variable $x$. The derivative of $x$ with respect to $x$ is equal to one as per the derivative rule of a variable.

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $x \times \dfrac{1}{x}$ $+$ $\log_{e}{x} \times 1$

It is time to simplify this differential equation. So, let’s focus on simplify the whole mathematical equation.

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $\dfrac{x \times 1}{x}$ $+$ $\log_{e}{x}$

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $\dfrac{x}{x}$ $+$ $\log_{e}{x}$

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $\dfrac{\cancel{x}}{\cancel{x}}$ $+$ $\log_{e}{x}$

$\implies$ $\dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $1+\log_{e}{x}$

$\implies$ $y \times \dfrac{1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

$\implies$ $\dfrac{y \times 1}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

$\implies$ $\dfrac{y}{y} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

$\implies$ $\dfrac{\cancel{y}}{\cancel{y}} \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

$\implies$ $1 \times \dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{(y)}$ $\,=\,$ $y \times (1+\log_{e}{x})$

The differential equation is obtained in terms of $x$ and $y$ but the exponential power function is given in terms of $x$ only. Hence, it is time to eliminate the variable $y$ by its assumed value. We have taken that the value of $y$ is equal to $x$ to the $x$-th power.

$\implies$ $\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ $\,=\,$ $x^{\displaystyle x} \times (1+\log_{e}{x})$

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ $\,=\,$ $x^{\displaystyle x}(1+\log_{e}{x})$

Thus, the differentiation of $x$ raised to the $x$-th power with respect to $x$ is evaluated in differential calculus by using the logarithms. It can also be written in the following form.

$\,\,\,\therefore\,\,\,\,\,\,$ $\dfrac{d}{dx}{\big(x^{\displaystyle x}\big)}$ $\,=\,$ $x^{\displaystyle x}(1+\ln{x})$

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